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An inductor of 10mH is connected to a 18V battery through a resistor of 10kOmegaand a switch. After a long time, when the maximum current is set up in the circuit, the current is switched off. Calculate the current in the circuit after 2 mus . |
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Answer» Solution :Given that `L = 10mH = 10^(-2) H, R= 10 k Omega = 10^4 Omega` and E = 18V , we KNOW that `I_0 = E/R= (18)/(10^4) A= 18 XX 10^(-4)A ` Time constant, `tau_L = L/R = (10 xx 10^(-3))/(10 xx 10^3) = 10^(-6)` sec We know that , `I= I_0 e^(-(R//L)t)` here `t = 2 mus = 2xx 10^(-6)S, I_0 = 18 xx 10^(-4)A` `I = 18 xx 10^(-4) = 2.48 xx 10^(-4)A` |
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